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Let $\zeta$ be a primitive $n^{\rm{th}}$ root of unity. Let $K=\mathbb{Q}(\zeta)$.

  • If $n=p^r (r\geq 1)$ is a prime power, show that $N_{K/F}(1-\zeta)=p$
  • If $n$ is divisible by at least two distinct primes then show that $N_{K/F}(1-\zeta)=1$

I tried for $r=1$

Suppose $n=p$ then we have $Gal(\mathbb{Q}(\zeta)/\mathbb{Q})\cong (\mathbb{Z}/p\mathbb{Z})^*=\{\sigma_1,\sigma_2,\sigma_3,\cdots,\sigma_{p-1}\}$ where $\sigma_i(\zeta)=\zeta^i$

Now,

$N_{K/F}(1-\zeta)=\sigma_1(1-\zeta)\sigma_2(1-\zeta)\cdots\sigma_{p-1}(1-\zeta)=(1-\zeta)(1-\zeta^2)\cdots(1-\zeta^{p-1})$

I could prove that this is $p$ by just expanding.. (as of now i do not have a general proof..)

When $r=2$ i tried to do similarly but ended up nowhere..

For $r=2$ we want to see what are all elements that are coprime to $p$.. (This is what gives me the galois group) Coprimes of $p^2$ are $\{1,2,\cdots, p^2\}-\{p,2p,3p,\cdots,p^2\}$.

Then, $N_{K/\mathbb{Q}}(1-\zeta)=(1-\zeta)(1-\zeta^2)\cdots(1-\zeta^{p-1})(1-\zeta^{p+1})(1-\zeta^{p+2})\cdots(1-\zeta^{p^2-1})$

This is equal to $$\frac{\Pi_{i=1}^{p^2-1}(1-\zeta^i)}{\Pi_{i=1}^{p-1}(1-\zeta^{ip})}$$

I do now know how to make this end up being $p$..

Please help me to clear this...

For $n=6$ we have $Gal(\mathbb{Q}(\zeta_6)/\mathbb{Q}=\{ \sigma_1,\sigma_5\}$

$N_{K/F}(1-\zeta)=(1-\zeta)(1-\zeta^5)=1-(\zeta+\zeta^5)+1=1$ (Credits to David Holden)

As i have succeeded in case of $n=6$ i tried to see this for $n=pq$ with $p,q$ are distinct primes..

co primes to $pq$ are $\{1,2,3,\cdots,pq-1\}-\{p,2p,3p,\cdots,(q-1)p, q,2q,3q,\cdots,(p-1)q\}$

Then $$N_{K/F}(1-\zeta)=\frac{\Pi_{i=1}^{pq-1}(1-\zeta^i)} {\Pi_{i=1}^{q-1}(1-\zeta^{ip})\Pi_{i=1}^{p-1}(1-\zeta^{iq})}$$

I do not know what should be done after this...

Please let me know how to fix with this...

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    $\begingroup$ $\zeta = e^{\frac{i\pi}3} = \frac12(1+\sqrt{3})$ so $\zeta +\zeta^5=\zeta+\zeta^{-1} = 1$ $\endgroup$ – David Holden Nov 1 '14 at 14:10
  • $\begingroup$ Oh... Yes.. @DavidHolden : I got it.... Thanks :) May be you should write $\frac{1}{2}(1+i\sqrt{3})$ $\endgroup$ – user87543 Nov 1 '14 at 14:27
  • $\begingroup$ yes, my mistake! $\endgroup$ – David Holden Nov 1 '14 at 14:30
  • $\begingroup$ @DavidHolden : That is alright :) :) Can you help me with this?? $\endgroup$ – user87543 Nov 1 '14 at 14:36
  • $\begingroup$ Is this a duplicate? Or this $\endgroup$ – Jyrki Lahtonen Nov 3 '14 at 21:03
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Extended hints for part (1):

  • The zeros of $f_r(x)=x^{p^r}-1$ are the numbers $\zeta^i, i=0,1,\ldots,p^r-1$.
  • Therefore the zeros of $f_r(x+1)=(x+1)^{p^r}-1$ are the numbers $\zeta^i-1, i=0,1,\ldots,p^r-1$. Notice that the constant term of this polynomial is equal to zero as $\zeta^0-1$ is one of the zeros.
  • Therefore the zeros of $g_r(x)=f_r(x+1)/x$ are the numbers $\zeta^i-1, i=1,2,\ldots,p^r-1$.
  • Therefore the constant term $g_r(0)$ is, up to sign, equal to the product $$g_r(0)=\pm\prod_{i=1}^{p^r-1}(1-\zeta^i).$$ You figure out the sign ;-)
  • Binomial theorem gives you the value of $g_r(0)$.
  • Doing this for both $r$ and $r-1$ should give you the answer.

For the second part the same idea should take you the distance. I haven't really thought about it more than that, so I'm only giving you a few suggestions. Let's denote $$ g_n(x)=\frac{(x+1)^n-1}x=x^{n-1}+\cdots+n. $$ Above I used $g_r$ instead of $g_{p^r}$. I apologize for changing my notation from part 1. The same calculation shows that $$ g_n(0)=\pm\prod_{k=1}^{n-1}(1-\zeta^k). $$ This time we have $$ N(1-\zeta)=\prod_{k=1, \gcd(k,n)=1}^{n-1}(1-\zeta^k). $$ So for example for if $n=pq$ is a product of two distinct primes, we get $$ \begin{aligned} N(1-\zeta)&=\frac{\prod_{k=1}^{n-1}(1-\zeta^k)} {\prod_{k=1,\gcd(k,pq)=p}^{n-1}(1-\zeta^k)\prod_{k=1,\gcd(k,pq)=q}^{n-1}(1-\zeta^k)}\\ &=\pm\frac{g_n(0)}{g_q(0)g_p(0)}. \end{aligned} $$ Again leaving it to you to figure out the signs.

At this point a bell rings. There is a well known formula for the cyclotomic polynomials $\Phi_n(x)$ as a fraction involving the polynomials $x^d-1$ with $d\mid n$. The Möbius function of $n/d$ tells you whether the factor $x^d-1$ goes upstairs, downstairs or is totally absent. I hazard a guess that you are expected to use that for the general case.

For example, if $n=pqr$ is a product of distinct primes $p,q,r$, then $$ \begin{aligned} N(1-\zeta)&=\prod_{k=1,\gcd(k,n)=1}^{n-1}(1-\zeta^k)\\ &=\frac{\left(\prod_{k=1}^{n-1}(1-\zeta^k)\right)\left(\prod_{k=1,pq\mid k}^{n-1}(1-\zeta^k)\right)\left(\prod_{k=1,pr\mid k}^{n-1}(1-\zeta^k)\right)\left(\prod_{k=1,rq\mid k}^{n-1}(1-\zeta^k)\right)} {\left(\prod_{k=1,p\mid k}(1-\zeta^k)\right)\left(\prod_{k=1,q\mid k}(1-\zeta^k)\right)\left(\prod_{k=1,r\mid k}(1-\zeta^k)\right)}\\ &=\frac{g_{pqr}(0)g_r(0)g_q(0)g_p(0)}{g_{rq}(0)g_{pr}(0)g_{pq}(0)} =\frac{pqr\cdot r\cdot q\cdot p}{rq\cdot pr\cdot pq}=1. \end{aligned} $$ The usual inclusion/exclusion business: We take all the powers. Next we cancel those with exponent of $\zeta$ divisible by $p$, $q$ or $r$. Then we notice that the exponents divisible by two prime factors where cancelled twice, so we need to add them there again ...

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  • $\begingroup$ Sir, as $\zeta^i-1$ are roots of $g(X)$ we have constant term $$g_r(0)=(-1)^{p^r-1}\prod_{i=1}^{p^r-1}(\zeta^i-1)$$ ass we want $(1-\zeta^i)$ we get another $(-1)^{p^r-1}$ so, we have $$g_r(0)=\prod_{i=1}^{p^r-1}(1-\zeta^i)$$ and that $g_r(0)=p^r$... So, we have $$p^r=\prod_{i=1}^{p^r-1}(1-\zeta^i)$$ But when we are taking norm we are not considering all powers of $\zeta$ but only certain powers of $\zeta$.. So, how should i proceed further? $\endgroup$ – user87543 Nov 3 '14 at 12:36
  • $\begingroup$ You already have the formula for the norm? The product of all $(1-\zeta^k)$ divided by a similar product of those with $p\mid k$. The latter product involves all the roots of unity of order that is a factor of $p^{r-1}$. Reread my last bullet. $\endgroup$ – Jyrki Lahtonen Nov 3 '14 at 13:20
  • $\begingroup$ I did not get your idea... You want me to know what $g_{r-1}(0)$ is.... :O I am lost! $\endgroup$ – user87543 Nov 3 '14 at 13:24
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    $\begingroup$ I want you to use the formula $$N(1-\zeta)=\frac{g_r(0)}{g_{r-1}(0)}.$$ I think that you have already proved that! $\endgroup$ – Jyrki Lahtonen Nov 3 '14 at 14:10
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    $\begingroup$ $$N(1-\zeta)=\prod_{k=1,\gcd(k,n)=1}^{n-1}(1-\zeta^k).$$ $$\Phi_n(x)=\prod_{k=1,\gcd(k,n)=1}^{n-1}(x-\zeta^k).$$ What do we get when we substitute $x=1$? I am very sorry that I didn't realize this much earlier :-( $\endgroup$ – Jyrki Lahtonen Nov 5 '14 at 5:28
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it seems that, in the case of a primitive $p$-th root, that the statement $N(1-\zeta)=p$ is equivalent to the assertion that: $$ \prod_{k=1}^{p-1} \sin(\frac{\pi k}{p}) = \frac{p}{2^{p-1}} $$

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I want to compute $$\frac{\prod_{i=1}^{p^2-1} (1-\zeta^i)}{\prod _{i=1}^{p-1}(1-\zeta^{ip})}$$

Consider $f(x)=x^{p^2}-1$.

Roots of $f(x)$ are $\zeta^{i}:0\leq i\leq p^2-1$ ($\zeta$ is primitive $p$th root of unity)

Roots of $f(x+1)=(x+1)^{p^2}$ are $\zeta^{i}-1:0\leq i\leq p^2-1$

Roots of $g(x)=\frac{f(x+1)}{x}$ are $\zeta^{i}-1:1\leq i\leq p^2-1$

Product of roots of $g(x)$ is $(-1)^{p^2-1}\prod_{i=1}^{p^2-1} (\zeta^i-1)=\prod_{i=1}^{p^2-1} (1-\zeta^i)$

By binomial expansion we see that constant term of $g(x)$ is $\binom{p^2}{p^2-1}=p^2$

Thus, $\prod_{i=1}^{p^2-1} (1-\zeta^i)=p^2$


Consider $h(x)=x^p-1$

Roots of $h(x)$ are $\eta^{i}:0\leq i\leq p-1$ ($\eta$ is primitive $p$th root of unity)

Roots of $h(x+1)=(x+1)^p$ are $\eta^{i}-1:0\leq i\leq p-1$

Roots of $p(x)=\frac{h(x+1)}{x}$ are $\zeta^{i}-1:1\leq i\leq p-1$

Product of roots of $p(x)$ is $(-1)^{p-1}\prod_{i=1}^{p^2-1} (\eta^i-1)=\prod_{i=1}^{p-1} (1-\eta^i)=\prod_{i=1}^{p-1} (1-\zeta^{ip})$

By binomial expansion we see that constant term of $g(x)$ is $\binom{p}{p-1}=p$

Thus, $\prod_{i=1}^{p^2-1} (1-\zeta^{ip})=p$.


So,$$\frac{\prod_{i=1}^{p^2-1} (1-\zeta^i)}{\prod _{i=1}^{p-1}(1-\zeta^{ip})}=p$$


More generally, $$\frac{\prod_{i=1}^{p^r-1} (1-\zeta^i)}{\prod _{i=1}^{p^{r-1}-1}(1-\zeta^{ip})}=p$$


For $n=pq$ we have

$$N_{K/F}(1-\zeta)=\frac{\prod_{i=1}^{pq-1}(1-\zeta^i)} {\prod_{i=1}^{q-1}(1-\zeta^{ip})\prod_{i=1}^{p-1}(1-\zeta^{iq})}$$

For similar reasons as above we have $\prod_{i=1}^{pq-1}(1-\zeta^i)=pq$ ; $\prod_{i=1}^{q-1}(1-\zeta^{ip}=q$ ; $\prod_{i=1}^{p-1}(1-\zeta^{iq}=p$

Thus, $$N_{K/F}(1-\zeta)=\frac{\prod_{i=1}^{pq-1}(1-\zeta^i)} {\prod_{i=1}^{q-1}(1-\zeta^{ip})\prod_{i=1}^{p-1}(1-\zeta^{iq})}=\frac{pq}{pq}=1$$

For $n=p^2q$ we have

$$N_{K/F}(1-\zeta)=\frac{\prod_{i=1}^{p^2q-1}(1-\zeta^i)\prod_{i=1}^{p-1}(1-\zeta^{iq})} {\prod_{i=1}^{pq-1}(1-\zeta^{ip})\prod_{i=1}^{p^2-1}(1-\zeta^{iq})}=\frac{p^2q\cdot p}{pq\cdot p^2}=1$$

Similar ideas for any $n$ having more than one prime in its prime factorization...

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We consider

$$\Phi_n(x)=\prod (x^{\frac{n}{d}}-1)^{\mu(d)}$$

If $n=p^2q^2$ then $$\Phi_{p^2q^2}(x)=(x^{p^2q^2}-1)(x^{p^2q}-1)^{-1}(x^{pq^2}-1)^{-1}(x^{pq}-1)=\frac{(x^{p^2q^2}-1)(x^{pq}-1)}{(x^{p^2q}-1)(x^{pq^2}-1)}=\frac{x^{p^2q-1}+x^{p^2q-2}+\cdots +x+1)(x^{pq-1}+x^{pq-2}+\cdots+x+1)}{(x^{p^2q-1}+x^{p^2q-2}+\cdots+x+1)(x^{pq^2-1}+x^{pq^2-2}+\cdots+x+1)}$$

Evaluating at $1$ we have :

$$\Phi_{p^2q^2}(1)=\frac{p^2q\cdot pq}{p^2q\cdot pq^2}=1$$

More generally, if $n=p_1^{i_1}p_2^{i_2}\cdots p_r^{i_r}$ (all $i_j$ are even, $r$ is also even) we have

$$\Phi_n(x)=\frac{(x^n-1)(x^{p_1p_2}-1)(x^{p_2p_3}-1)\cdots ??}{???}$$

I am not able to write general expression of $\Phi_n(x)$ neatly but $\Phi_n(1)$ is $$\frac{p_1^{i_1}p_2^{i_2}\cdots p_r^{i_r}\cdot(p_1p_2)(p_2p_3)\cdots (p_1p_2p_3p_4)\cdots}{:( ~:(}$$


We consider another formula:

If $p$ is a prime number that does not divide $n$ then we have

$$\Phi_{pn}(x)=\frac{\Phi_n(x^p)}{\Phi_n(x)}$$

If $p$ divides $n$ then we have $\Phi_{pn}(x)=\phi_n(x^p)$

For $n=pq$ we have $$\Phi_{pq}(x)=\frac{\Phi_q(x^p)}{\Phi_q(x)}\Rightarrow \Phi_{pq}(1)=\frac{\Phi_q(1)}{\Phi_q(1)}=\frac{q}{q}=1$$

For $n=pqr$ we have $$\Phi_{pqr}(x)=\frac{\Phi_{pq}(x^r)}{\Phi_{pq}(x)}\Rightarrow \Phi_{pqr}(1)=\frac{\Phi_{pq}(1)}{\Phi_{pq}(1)}=\frac{1}{1}=1$$

More generally, for $n=p_1p_2p_\cdots p_r $ we have $\Phi_n(1)=1$

Even more generally, for $n=p_1^{i_1}p_2^{i_2}\cdots p_r^{i_r}$ we have

$$\Phi_{p_1^{i_1}p_2^{i_2}\cdots p_r^{i_r}}(x)=\cdots = \Phi_{p_1p_2\cdots p_r} (x^{p_1^{i_1-1}p_2^{i_2-1}\cdots p_r^{i_r-1}})$$

So, $$\Phi_n(1)=\Phi_{p_1^{i_1}p_2^{i_2}\cdots p_r^{i_r}}(1)=\cdots = \Phi_{p_1p_2\cdots p_r} (1)=1$$

Assuming $N_{K/\mathbb{Q}}(1-\zeta)=\Phi_n(1)$ we conclude that , if $n$ has atleast two distinct primes in its factorization then $$N_{K/\mathbb{Q}}(1-\zeta)=\Phi_n(1)=1$$

If $n=p$ we have $$\Phi_p(x)=x^{p-1}+x^{p-2}+\cdots +x+1\Rightarrow N_{K/\mathbb{Q}}(1-\zeta)=\Phi_p(1)=p$$

If $n=p^r$ then,

$$\Phi_n(x)=\Phi_p(x^{p^{r-1}})\Rightarrow \Phi_n(1)=\Phi_p(1)\Rightarrow N_{K/\mathbb{Q}}(1-\zeta)=\Phi_n(1)=\Phi_p(1)=p$$

So, we have

  • $N_{K/\mathbb{Q}}(1-\zeta)=p$ if $n=p^r$
  • $N_{K/\mathbb{Q}}(1-\zeta)=1$ if $n$ is divisible by at least two distinct primes
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